Explanation Write the expression for dryness fraction at point 2s. x 2 s = s 2 s − s f 2 s g 2 − s f 2 (I) Here, specific entropy of saturated liquid at state 2 is s f 2 , specific entropy of saturated vapor at state 2 is s g 2 and the dryness fraction ay point 2s is x 2 s . Write the expression for specific enthalpy at point 2s. h 2 s = h f 2 + x 2 s ( h g 2 − h f 2 ) . (II) Here, specific enthalpy of saturated liquid at state 2 is h f 2 , specific enthalpy of saturated vapor at state 2 is h g 2 and the specific enthalpy at state 2s is h 2 s . Write the expression for power developed by the turbine. W ˙ c v = m ˙ η t ( h 1 − h 2 ) (III) Here, power developed by the turbine is W ˙ c v , mass flow rate is m ˙ , efficiency of the turbine is η t , specific enthalpy at state2 is h 2 and the specific enthalpy at state1 is h 1 . Conclusion: Refer to Table A-4, “Properties of superheated water vapor” at 0.1 bar to obtain the value of s g 2 as 8.1502 kJ / kg ⋅ K , s f 2 as 0.6493 kJ / kg ⋅ K , h f 2 as 191.83 kJ / kg ⋅ K and h g 2 as 258.47 kJ / kg ⋅ K . Substitute 6.320 kJ / kg ⋅ K for s 2 s , 8.1502 kJ / kg ⋅ K for s g 2 and 0.6493 kJ / kg ⋅ K for s f 2 in equation (I). x 2 s = 6.320 kJ / kg ⋅ K − 0.6493 kJ / kg ⋅ K 8.1502 kJ / kg ⋅ K − 0.6493 kJ / kg ⋅ K = 5.6707 kJ / kg ⋅ K 7.5009 kJ / kg ⋅ K = 0.756 Substitute 2584.7 kJ / kg ⋅ K for h g 2 , 191.83 kJ / kg ⋅ K for h f 2 and 0.756 for x 2 s in equation (II)

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ISBN: 9781119391630

Author: MORAN

Publisher: WILEY

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Chapter 6.13, Problem 144P

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Chapter 6.13, Problem 143P

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